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Problem 6.31
A steel bar has a rectangular cross setion 12.7 mm wide and 38.1 mm deep.
The bar is subjected to a
twisting moment
T
=135.7 Nm.
The shear yield stress of the material is 82.7 MPa.
(a) By Eq. 6.64, calculate the maximum shear stress in the bar and show in a diagram where it occurs.
(b) Calculate the shear stress in the bar at the center of the short side.
Solution:
Location of
τ
max
(a)
τ
max
T
k
2
2b
⋅
()
⋅
2h
⋅
2
⋅
=
Eq. 6.64
T
135.7 N
⋅
m
⋅
:=
h
12.7 mm
⋅
2
:=
b
38.1 mm
⋅
2
:=
b
h
3
=
From Table 6.1, page 225
k
1
0.263
:=
k
2
0.267
:=
τ
max
T
k
2
⋅
⋅
⋅
2
⋅
82.706 MPa
=
:=
(b) To obtain the shear stress at the center of the short side,
interchange (2b) and (2h) in Eq. 6.64)
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This note was uploaded on 01/11/2011 for the course MAE 3040 taught by Professor Mechanicsofsolids during the Spring '10 term at Utah State University.
 Spring '10
 MechanicsofSolids
 Shear, Stress

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